What will be the output of the program ?

#include<stdio.h>

int main()
{
    void fun(int, int[]);
    int arr[] = {1, 2, 3, 4};
    int i;
    fun(4, arr);
    for(i=0; i<4; i++)
        printf('%d,', arr[i]);
    return 0;
}
void fun(int n, int arr[])
{
    int *p=0;
    int i=0;
    while(i++ < n)
        p = &arr[i];
    *p=0;
}

ANS:A - 2, 3, 4, 5

Step 1: void fun(int, int[]); This prototype tells the compiler that the function fun() accepts one integer value and one array as an arguments and does not return anything. Step 2: int arr[] = {1, 2, 3, 4}; The variable a is declared as an integer array and it is initialized to a[0] = 1, a[1] = 2, a[2] = 3, a[3] = 4 Step 3: int i; The variable i is declared as an integer type. Step 4: fun(4, arr); This function does not affect the output of the program. Let's skip this function. Step 5: for(i=0; i<4; i++) { printf('%d,', arr[i]); } The for loop runs untill the variable i is less than '4' and it prints the each value of array a. Hence the output of the program is 1,2,3,4